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Chapter 40: Problem 56

(III) A \(120-\mathrm{V}\) rms \(60-\mathrm{Hz}\) voltage is to be rectified with a full-wave rectifier as in Fig. \(40,\) where \(R=28 \mathrm{k} \Omega\) , and \(C=35 \mu \mathrm{F}\) (a) Make a rough estimate of the average current. \((b)\) What happens if \(C=0.10 \mu \mathrm{F} ?\)

Short Answer

Expert verified
Average current with \(C=35 \mu\text{F}\) is approximately \(3.857 \, \text{mA}\), and it decreases if \(C=0.10 \mu\text{F}\) due to a higher ripple.

Step by step solution

01

Rectifier Output Voltage (V_peak)

First, calculate the peak voltage \( V_{\text{peak}} \) from the given rms voltage using the formula:\[V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2}\]Given \( V_{\text{rms}} = 120 \text{ V} \), so\[V_{\text{peak}} = 120 \times \sqrt{2} \approx 169.71 \text{ V}\]
02

Average Voltage (V_avg)

For a full-wave rectifier, the average output voltage \( V_{\text{avg}} \) is \( \frac{2}{\pi} \) of the peak voltage:\[V_{\text{avg}} = \frac{2}{\pi} V_{\text{peak}} = \frac{2}{\pi} \times 169.71 \approx 108.00 \text{ V}\]
03

Calculate Average Current (I_avg)

Use Ohm's Law to find the average current \( I_{\text{avg}} \) with the calculated average voltage and the resistance \( R \):\[I_{\text{avg}} = \frac{V_{\text{avg}}}{R} = \frac{108.00}{28000} \approx 0.003857 \text{ A} \, (3.857 \, \text{mA})\]
04

Effect of Changing Capacitance (C)

When \( C = 0.10 \mu \text{F} \), the ripple voltage increases because there is less capacitance to smooth the output. As a result, the average voltage might decrease and, consequently, the average current would also decrease. However, calculating precise effects would require additional steps involving the ripple factor calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rectifier Output Voltage
In a full-wave rectifier circuit, the rectifier output voltage is a crucial part to understand. The first thing you need to calculate is the peak voltage (\( V_{\text{peak}} \)), which determines the maximum voltage reached in each cycle of the AC input signal. For this, the formula is:
  • \[V_{ ext{peak}} = V_{ ext{rms}} \times \sqrt{2}\]
With a given rms voltage (\( V_{\text{rms}} \)) of 120 V, you simply multiply it by the square root of 2. This calculation reveals that the peak voltage is approximately 169.71 V. This represents the highest point of voltage that your rectified signal will achieve.
The rectified output voltage is now ready to be averaged across the output signal to find its effective value.
Average Current Calculation
Average current calculation in a full-wave rectifier system relies on understanding both the average voltage coming out of the rectifier and the resistive load it handles. For a full-wave rectifier, the average output voltage (\( V_{\text{avg}} \)) is calculated as:
  • \[V_{ ext{avg}} = \frac{2}{\pi} V_{ ext{peak}}\]
Here, with \( V_{\text{peak}} = 169.71 \; \text{V} \), the result is an average voltage of about 108 V.
Next, to find the average current (\( I_{\text{avg}} \)), you apply Ohm's Law, which is:
  • \[I_{ ext{avg}} = \frac{V_{ ext{avg}}}{R}\]
Using a resistance (\( R \)) of 28 k惟, the average current comes out to approximately 3.857 mA. This current flows continuously in the rectifier's DC load, an important factor when considering power and efficiency.
Effect of Capacitance on Ripple Voltage
Capacitance plays a crucial role in smoothing out the ripple voltage in rectified voltage. When the capacitance is high, it can charge and discharge to stabilize the voltage more effectively. When the capacitance is low, however, the circuit cannot smooth the output as well.
For instance, with a significant capacitance of 35 碌F, the ripple voltage in the system will be minimal, giving you a more stable DC output. However, if this capacitance is reduced to 0.10 碌F, the ripple voltage increases substantially. This increase means you have less smoothness in your output waveform, potentially decreasing your average voltage and, consequently, your average current.
Thus, when designing circuits, one must carefully balance capacitance to meet the required levels of voltage stability and performance.

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